POJ3026-Borg Maze(廣度優先搜索+最小生成樹)

Borg Maze

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12346		Accepted: 4046

Description

The Borg is an immensely powerful race of enhanced humanoids from the delta quadrant of the galaxy. The Borg collective is the term used to describe the group consciousness of the Borg civilization. Each Borg individual is linked to the collective by a sophisticated subspace network that insures each member is given constant supervision and guidance. 

Your task is to help the Borg (yes, really) by developing a program which helps the Borg to estimate the minimal cost of scanning a maze for the assimilation of aliens hiding in the maze, by moving in north, west, east, and south steps. The tricky thing is that the beginning of the search is conducted by a large group of over 100 individuals. Whenever an alien is assimilated, or at the beginning of the search, the group may split in two or more groups (but their consciousness is still collective.). The cost of searching a maze is definied as the total distance covered by all the groups involved in the search together. That is, if the original group walks five steps, then splits into two groups each walking three steps, the total distance is 11=5+3+3.

Input

On the first line of input there is one integer, N <= 50, giving the number of test cases in the input. Each test case starts with a line containg two integers x, y such that 1 <= x,y <= 50. After this, y lines follow, each which x characters. For each character, a space `` '' stands for an open space, a hash mark ``#'' stands for an obstructing wall, the capital letter ``A'' stand for an alien, and the capital letter ``S'' stands for the start of the search. The perimeter of the maze is always closed, i.e., there is no way to get out from the coordinate of the ``S''. At most 100 aliens are present in the maze, and everyone is reachable.

Output

For every test case, output one line containing the minimal cost of a succesful search of the maze leaving no aliens alive.

Sample Input

2
6 5
##### 
#A#A##
# # A#
#S  ##
##### 
7 7
#####  
#AAA###
#    A#
# S ###
#     #
#AAA###
#####  

Sample Output

8
11

題目大意

在一個y行 x列的迷宮中，有可行走的通路空格’ ‘，不可行走的牆’#’，還有兩種英文字母A和S，現在從S出發，要求用最短的路徑L連接所有字母，輸出這條路徑L的總長度。

題意明白之後就是解題方法了，最短的路徑我們可以通過BFS找出，最小生成樹可以使用Prim

坑點是數據的任意一行的最後都有一個空格，所以scanf("%d\n");的時候一定要加一個\n將多餘的空格過濾掉，字符串就用gets就可以了。

#include <algorithm>
#include <iostream>
#include <numeric>
#include <cstring>
#include <iomanip>
#include <string>
#include <vector>
#include <cstdio>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#define LL long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const LL Max = 1005;
const double esp = 1e-6;
//const double PI = acos(-1);
const int INF = 0x3f3f3f3f;
using namespace std;

char arr[60][60];
bool vis[60][60];
struct node{
    int x,y,z;
}cnt,ant;
int Map[105][105],n,m,top;
int dis[105],Pos[60][60];
bool Vis[105];
int dx[] = {0,0,1,-1};
int dy[] = {1,-1,0,0};

void cal(int x,int y){ //搜索找出所有字母之間的最短距離
    memset(vis,false,sizeof(vis));
    queue<node>Q;
    ant.x = x;ant.y = y;ant.z = 0;
    vis[ant.x][ant.y] = true;
    Q.push(ant);
    while(!Q.empty()){
        ant = Q.front();Q.pop();
        if(arr[ant.x][ant.y] == 'A' || arr[ant.x][ant.y] == 'S'){
            Map[Pos[x][y]][Pos[ant.x][ant.y]] = Map[Pos[ant.x][ant.y]][Pos[x][y]] = ant.z;
        }
        for(int i=0;i<4;i++){
            int nowx = ant.x + dx[i];
            int nowy = ant.y + dy[i];
            if(nowx >= 0 && nowx < m && nowy >= 0 && nowy < n && arr[nowx][nowy]!='#' && !vis[nowx][nowy]){
                vis[nowx][nowy] = true;
                cnt.x = nowx;cnt.y = nowy;cnt.z = ant.z + 1;
                Q.push(cnt);
            }
        }
    }
}
int Prim(int x){//最小生成樹，找出所有節點之間的總長度
    memset(Vis,false,sizeof(Vis));
    for(int i=1;i<top;i++)
        dis[i] = Map[1][i];
    int ans = 0;
    Vis[x] = true;
    for(int i=2;i<top;i++){
        int qmin = INF,k = i;
        for(int j=1;j<top;j++){
            if(!Vis[j] && dis[j] < qmin){
                qmin = dis[j];
                k = j;
            }
        }
        ans += qmin;
        Vis[k] = true;
        for(int j=1;j<top;j++){
            if(!Vis[j] && dis[j] > Map[k][j]){
                dis[j] = Map[k][j];
            }
        }
    }
    return ans;
}
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d %d\n",&n,&m);
        top = 1;
        for(int i=0;i<m;i++){
            gets(arr[i]);
            for(int j=0;j<n;j++){
                if(arr[i][j] == 'S' || arr[i][j] == 'A'){
                    Pos[i][j] = top;
                    top += 1;
                }
            }
        }
        for(int i=0;i<=top;i++)
            for(int j=0;j<=top;j++)
                if(i == j)
                    Map[i][j] = 0;
                else
                    Map[i][j] = INF;
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(arr[i][j] == 'S' || arr[i][j] == 'A'){
                    cal(i,j);
                }
            }
        }
        printf("%d\n",Prim(1));
    }
    return 0;
}